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Placement Papers>>Wipro

# Wipro InfoTech Placement Papers

Wipro paper(System software)
July-1997
------------

PART --A
------------------------------------------------------
1) abcD+abcd+aBCd+aBCD
then the simplified function is
( Capital letters are copliments of corresponding letters
A=compliment of a)

[a] a [b] ab [c] abc [d] a(bc)* [e] mone
(bc)*=compliment of bc

Ans: e

-------------------------------------
2) A 12 address lines maps to the memory of

[a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none

Ans: b

----------------------------------------
3) In a processor these are 120 instructions . Bits needed to impliment
this instructions
[a] 6 [b] 7 [c] 10 [d] none

Ans: b

-----------------------------------------
4) In 8085 microprocessor READY signal does.which of the following
is incorrect statements
[a]It is input to the microprocessor
[b] It sequences the instructions

Ans : b
----------------------------------------
--More--
----------------------------------------

5) Return address will be returned by function to
[a] Pushes to the stack by call
Ans : a
------------------------------------------
6)

n=7623
{
temp=n/10;
result=temp*10+ result;
n=n/10
}

Ans : 3267
----------------------------------------------
7) If A>B then
F=F(G);
else B>C then
F=G(G);
in this , for 75% times A>B and 25% times B>C then,is 10000 instructions
are there ,then the ratio of F to G
[a] 7500:2500 [b] 7500:625 [c] 7500:625 if a=b=c else
7500:2500
--------------------------------------------------
8) In a compiler there is 36 bit for a word and to store a character 8bits are
needed. IN this to store
a character two words are appended .Then for storing a K characters string,
How many words are needed.
[a] 2k/9 [b] (2k+8)/9 [c] (k+8)/9 [d] 2*(k+8)/9 [e] none

Ans: a
---------------------------------------------------------
9) C program code

int zap(int n)
{
if(n<=1)then zap=1;
--More--
else zap=zap(n-3)+zap(n-1);
}
then the call zap(6) gives the values of zap
[a] 8 [b] 9 [c] 6 [d] 12 [e] 15

Ans: b
---------------------------------------------------------------

PART-B
-------
1) Virtual memory size depends on

[a] address lines [b] data bus
[c] disc space [d] a & c [e] none

Ans : a
-----------------------------------------------
2) Critical section is
[a]
[b] statements which are accessing shared resourses
Ans : b
-------------------------------------------------

mul a
store t1
mul b
store t2
mul t2

then the content in accumulator is

Ans : a**2+b**4
---------------------------------------------------
4) question (3) in old paper
--More--
5) q(4) in old paper
6) question (7) in old paper
7) q(9) in old paper
------------------------------

Hughes,Delhi:
-------------

> > (A)Aptitude :25 Qns, 20 Minutes
> >
> > 1. 2 x 4 analytical GRE type qns
> > 2. 2-3 Reasoning qns (GRE type)
> > 3. Probability of getting a sum of 7 when two dices are thrown together
> > 4. Rest quantitative questions
> >
> > (B) Technical: 50 Qns, 45 Minutes
> >
> > 1. 3 qns on operating systems. I qn on dijkestra algorithm

1. 3 qns on operating systems. I qn on dijkestra algorithm
> >
> > 2. Using which pin it's possible to address 16 bit addresses even though
there
> > are only 8 address bits in 8085? Ans: ALE
> > 3. Voltage gain for an amplifier is 100 while it is operating at 10
volts.
> > What is the O/P voltage wen i/p is 1 volt
> > 4. Quality factor indicates a0 Quality of inductor b) quality of
capacitor
> > c) both
> > 5. Qns related to bridges, routers and generators, which OSI layer they
> > corresspond to. (Refer to stevens 4th chapter)
> > 6.OPAmp's I/P ciurrent, O/p current and CMRR is given, what is the
voltage
> > gain
> > 7. 2-3 qns on scope of static variables in C. Qn to view o/p odf a C
static
> > var
> > 3. Voltage gain for an amplifier is 100 while it is operating at 10
volts.
> > What is the O/P voltage wen i/p is 1 volt
> > 4. Quality factor indicates a0 Quality of inductor b) quality of
capacitor
> > c) both
> > 5. Qns related to bridges, routers and generators, which OSI layer they
> > corresspond to. (Refer to stevens 4th chapter)
> > 6.OPAmp's I/P ciurrent, O/p current and CMRR is given, what is the
voltage
> > gain
> > 7. 2-3 qns on scope of static variables in C. Qn to view o/p odf a C
static
> > var
of static variables in C. Qn to view o/p odf a C static
> > var
> > 8. Qn to print a value of a pointer
> > 9.resistance increases with temperature in a) Metal b) semiconductor
> > 10. A qn to find the physical address from a given virtual address,
virtual
> > to physical address table was provided
> > 11. 16 bit mantissa and 8 bit exponent can present what maximum value?
> > 12. 4 bit window size in sliding window protocol, how many
acknowledements can be held?
> > 13. Security functionality is provided by which layer of OSI
> > 14. Frequency spectrums for AM, FM and PM (figure given, u'veto tell
which
> > Kind of modulation it belongs to)
> > 15. Among AM and FM which is better and why?
> > 16.LASt stage of TTL NAND gate is called: Ans: Totem Pole Amplifier
> > 17. SR to JK flip flop conversion. Ans: S=JQ', R=KQ
> > 18. LSB of a shift register is connected to its MSB, what is formed:
Ans:
> > RING Counter
> > 19. 2-3 Qns based on Demorgan's laws (identiies: (A+b)' = A'b', etc)
> > 20. 2 qns on Logic gates (O/p of logic gates)
> > 21. Diff in IRET and RET statements of 8086
> > 22. How many address bytes are required to address an array of memory
chips
> > (4 * 6), each chip having 4 memory bits and 8k registers.
> > 23. Diff. in memory mapped and I/P O/P mapped Input/Output (Refer a book

23. Diff. in memory mapped and I/P O/P mapped Input/Output (Refer a book
on Microprocessor)
> > 24. Qn on pipeline architecture
> > 25 QN on LAPB protocol
> >

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**************************************************************************

WIPRO INFOTECH

------------------------------------------------------------------------------
Date: July 1997
source: Roorkee University
------------------------------------------------------------------------------
This paper have two sections
each paper contains 10 questions each.
-----------------------------------------------------------------------------
1) abcD+abcd+aBCd+aBCD
then the simplified function is
( Capital letters are copliments of corresponding letters
A=compliment of a)

[a] a [b] ab [c] abc [d] a(bc)* [e] mone
(bc)*=compliment of bc

Ans: e

-------------------------------------
2) A 12 address lines maps to the memory of

[a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none

Ans: b

----------------------------------------
3) In a processor these are 120 instructions . Bits needed to impliment
this instructions
[a] 6 [b] 7 [c] 10 [d] none

Ans: b

-----------------------------------------
4) In 8085 microprocessor READY signal does.which of the following
is incorrect statements
[a]It is input to the microprocessor
[b] It sequences the instructions

Ans : b
----------------------------------------

5) Return address will be returned by function to
[a] Pushes to the stack by call
Ans : a
------------------------------------------
6)
n=7623
{
temp=n/10;
result=temp*10+ result;
n=n/10
}

Ans : 3267
----------------------------------------------
7) If A>B then
F=F(G);
else B>C then
F=G(G);
in this , for 75% times A>B and 25% times B>C then,is 10000 instructions
are there ,then the ratio of F to G
[a] 7500:2500 [b] 7500:625 [c] 7500:625 if a=b=c else
7500:2500
--------------------------------------------------
8) In a compiler there is 36 bit for a word and to store a character 8bits are needed. IN this to store
a character two words are appended .Then for storing a K characters string,
How many words are needed.
[a] 2k/9 [b] (2k+8)/9 [c] (k+8)/9 [d] 2*(k+8)/9 [e] none

Ans: a
---------------------------------------------------------
9) C program code

int zap(int n)
{
if(n<=1)then zap=1;
else zap=zap(n-3)+zap(n-1);
}
then the call zap(6) gives the values of zap
[a] 8 [b] 9 [c] 6 [d] 12 [e] 15

Ans: b
---------------------------------------------------------------

PART-B
-------
1) Virtual memory size depends on
[a] address lines [b] data bus
[c] disc space [d] a & c [e] none

Ans : a
-----------------------------------------------
2) Critical section is
[a]
[b] statements which are accessing shared resourses
Ans : b
-------------------------------------------------

mul a
store t1
mul b
store t2
mul t2

then the content in accumulator is

Ans : a**2+b**4
---------------------------------------------------
4) question (3) in old paper
5) q(4) in old paper
6) question (7) in old paper
7) q(9) in old paper
------------------------------

> Wipro paper(System software)
> July-1997
> ------------
>
> PART --A
> ------------------------------------------------------
> 1) abcD+abcd+aBCd+aBCD
> then the simplified function is
> ( Capital letters are copliments of corresponding letters
> A=compliment of a)
>
> [a] a [b] ab [c] abc [d] a(bc)* [e] mone
> (bc)*=compliment of bc
>
> Ans: e
>
> -------------------------------------
> 2) A 12 address lines maps to the memory of
>
> [a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none
>
> Ans: b
>
> ----------------------------------------
> 3) In a processor these are 120 instructions . Bits needed to impliment
> this instructions
> [a] 6 [b] 7 [c] 10 [d] none
>
> Ans: b
>
> -----------------------------------------
> 4) In 8085 microprocessor READY signal does.which of the following
> is incorrect statements
> [a]It is input to the microprocessor
> [b] It sequences the instructions
>
> Ans : b
> ----------------------------------------
>
> 5) Return address will be returned by function to
> [a] Pushes to the stack by call
> Ans : a
> ------------------------------------------
> 6)
> n=7623
> {
> temp=n/10;
> result=temp*10+ result;
> n=n/10
> }
>
> Ans : 3267
> ----------------------------------------------
> 7) If A>B then
> F=F(G);
> else B>C then
> F=G(G);
> in this , for 75% times A>B and 25% times B>C then,is 10000 instructions
> are there ,then the ratio of F to G
> [a] 7500:2500 [b] 7500:625 [c] 7500:625 if a=b=c else
> 7500:2500
> --------------------------------------------------
> 8) In a compiler there is 36 bit for a word and to store a character 8bits are
> needed. IN this to store
> a character two words are appended .Then for storing a K characters string,
> How many words are needed.
> [a] 2k/9 [b] (2k+8)/9 [c] (k+8)/9 [d] 2*(k+8)/9 [e] none
>
> Ans: a
> ---------------------------------------------------------
> 9) C program code
>
> int zap(int n)
> {
> if(n<=1)then zap=1;
> else zap=zap(n-3)+zap(n-1);
> }
> then the call zap(6) gives the values of zap
> [a] 8 [b] 9 [c] 6 [d] 12 [e] 15
>
> Ans: b
> ---------------------------------------------------------------
>
>
> PART-B
> -------
> 1) Virtual memory size depends on
> [a] address lines [b] data bus
> [c] disc space [d] a & c [e] none
>
> Ans : a
> -----------------------------------------------
> 2) Critical section is
> [a]
> [b] statements which are accessing shared resourses
> Ans : b
> -------------------------------------------------
>
> mul a
> store t1
> mul b
> store t2
> mul t2
>
> then the content in accumulator is
>
> Ans : a**2+b**4
> ---------------------------------------------------
> 4) question (3) in old paper
> 5) q(4) in old paper
> 6) question (7) in old paper
> 7) q(9) in old paper
> ------------------------------

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